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# Exercise 50 - Chapter 1

edited April 2018

We can make any set $$X$$ into a poset where we say $$x \leq y$$ if and only if $$x = y$$. A poset of this form is called the discrete poset.

Show that monotone maps between discrete posets are just functions.

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1.

Monotone maps are functions by definition. I'm not sure what this exercise was really asking.

Comment Source:Monotone maps are functions by definition. I'm not sure what this exercise was really asking.
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2.
edited April 2018

I think the exercise is getting at the issue that there is no monotonic relation that needs to be preserved by the mapping, hence it is "just" a function.

Comment Source:I think the exercise is getting at the issue that there is no monotonic relation that needs to be preserved by the mapping, hence it is "just" a function.
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3.
edited April 2018

Let $$P, Q$$ be discrete preorders, and let $$f : P \to Q$$ be a function. Let $$x, y \in P$$, and suppose $$x \le y$$; then $$x = y$$, since $$P$$ is a discrete preorder. So $$f(x) = f(y)$$, hence $$f(x) \le f(y)$$. Therefore, $$f$$ is a monotone map.

Notice that we didn't actually use the fact that $$Q$$ is a discrete preorder. Thus, any function out of a discrete preorder (to an arbitrary preorder) is a monotone map.

Comment Source:Let \$$P, Q\$$ be discrete preorders, and let \$$f : P \to Q\$$ be a function. Let \$$x, y \in P\$$, and suppose \$$x \le y\$$; then \$$x = y\$$, since \$$P\$$ is a discrete preorder. So \$$f(x) = f(y)\$$, hence \$$f(x) \le f(y)\$$. Therefore, \$$f\$$ is a monotone map. Notice that we didn't actually use the fact that \$$Q\$$ is a discrete preorder. Thus, any function out of a discrete preorder (to an arbitrary preorder) is a monotone map.
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4.
edited April 2018

When a mathematician says "monotone maps between discrete posets are just functions", they mean that every function between discrete posets is a monotone map.

This is much stronger than saying "monotone maps between discrete posets are functions", which is also true, but boring. The word just changes everything.

But Jonathan is right: we can say even more. Remember, for any set $$X$$:

1. Setting $$x \leq y$$ if and only if $$x = y$$ gives you a preorder on $$X$$ called the discrete preorder.

2. Setting $$x \leq y$$ for all $$x,y \in X$$ gives you a preorder on $$X$$ called the codiscrete preorder.

Any function from a discrete preorder to any preorder is a monotone map. Similarly, any function from any preorder to a codiscrete preorder is a monotone map!

Comment Source:When a mathematician says "monotone maps between discrete posets are _just_ functions", they mean that _every_ function between discrete posets is a monotone map. This is much stronger than saying "monotone maps between discrete posets are functions", which is also true, but boring. The word _just_ changes everything. But Jonathan is right: we can say even more. Remember, for any set \$$X\$$: 1. Setting \$$x \leq y\$$ if and only if \$$x = y\$$ gives you a preorder on \$$X\$$ called the **discrete** preorder. 2. Setting \$$x \leq y \$$ for all \$$x,y \in X\$$ gives you a preorder on \$$X\$$ called the **codiscrete** preorder. Any function from a discrete preorder to any preorder is a monotone map. Similarly, any function from any preorder to a codiscrete preorder is a monotone map!
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5.

John wrote:

Setting $$x \leq y$$ for all $$x,y \in X$$ gives you a preorder on $$X$$ called the codiscrete preorder.

Why is this preorder called "codiscrete"? In what sense is it dual to the discrete preorder?

Usually to obtain the dual of a category we reverse its arrows; but for the discrete preorder, reversing the inequalities should still give a discrete preorder.

Comment Source:[John](https://forum.azimuthproject.org/profile/17/John%20Baez) wrote: > Setting \$$x \leq y \$$ for all \$$x,y \in X\$$ gives you a preorder on \$$X\$$ called the **codiscrete** preorder. Why is this preorder called "codiscrete"? In what sense is it dual to the discrete preorder? Usually to obtain the dual of a category we reverse its arrows; but for the discrete preorder, reversing the inequalities should still give a discrete preorder. 
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6.

I assume it has something to do with this:

Any function from a discrete preorder to any preorder is a monotone map. Similarly, any function from any preorder to a codiscrete preorder is a monotone map!

I'd love to hear what the reason is well!

Comment Source:I assume it has something to do with this: > Any function from a discrete preorder to any preorder is a monotone map. Similarly, any function from any preorder to a codiscrete preorder is a monotone map! I'd love to hear what the reason is well!